In right triangle $MNO$, $\tan{M}=\frac{5}{4}$, $OM=8$, and $\angle O = 90^\circ$.  Find $MN$.  Express your answer in simplest radical form.
Solution: Our triangle is as shown below.

[asy]
size(100);
draw((0,0)--(8,0)--(0,10)--cycle,black+linewidth(1));
draw(rightanglemark((8,0),(0,0),(0,10),20),black+linewidth(1));
label("$O$",(0,0),W);
label("$M$",(8,0),E);
label("$N$",(0,10),W);
label("8",(0,0)--(8,0),S);
[/asy]

Since $\tan{M}=\frac{5}{4}$, we have $\dfrac{NO}{OM} = \dfrac{5}{4}$, so  $$NO = \frac{5}{4}OM = \frac{5}{4}\cdot 8 = 10.$$Then, from the Pythagorean Theorem, we have \begin{align*}
MN&=\sqrt{NO^2+OM^2}\\
&=\sqrt{10^2+8^2}=\sqrt{164}=\boxed{2\sqrt{41}}.\end{align*}